$\lim\limits_{x\rightarrow+\infty}\left[(x^3+x^2+1)^{1/3}- (x^2+x+1)^{0.5}\right] = -1/6$? Every method I use seem to get me to something to the extent of $0/0$, stuff I can't work with. Wolfram Alpha claims the answer to this is $-1/6$ but they offer no step by step solution.
Would appreciate any tips and help.
 A: Since $$\alpha-\beta=\frac{\alpha^6-\beta^6}{\alpha^5+...+\beta^5}$$
we have that
\begin{gather}\lim\limits_{x\rightarrow+\infty}({x^3+x^2+1})^{1/3}-{(x^2+x+1)}^{1/2}=\lim\limits_{x\rightarrow+\infty}x[{(1+1/x+1/{x^2})}^{1/3}-{(1+1/x+1/{x^2})^{1/2}}]=
\lim\limits_{x\rightarrow+\infty}x\frac{(1+1/x+1/{x^2})^2-(1+1/x+1/{x^2})^3}{(1+1/x+1/{x^2})^{5/3}+...+{(1+1/x+1/{x^2})}^{5/2}}
\end{gather}
The denominator goes to  $6$ and the numerator:
\begin{gather}\lim\limits_{x\rightarrow+\infty}x[(1+1/x+1/{x^2})^2-(1+1/x+1/{x^2})^3]=\\
\lim\limits_{x\rightarrow+\infty}x(1+1/x+1/{x^2})^2[-1/x-1/{x^2}]=\\
\lim\limits_{x\rightarrow+\infty}-(1+1/x+1/{x^2})^2[1+1/x]=-1
\end{gather}
A: $$  (x^3+x^2+1)^{1/3} - (x^2+x+1)^{0.5} = x \cdot (1+x^{-1}+x^{-3})^{1/3} - x \cdot (1+x^{-1}+x^{-2})^{0.5} $$
then I suggest Taylor series for both terms. 
$$ x \cdot \left [ 1+\frac{1}{3} \cdot x^{-1}-\frac{1}{9} \cdot x^{-2} + ... - 1-\frac{1}{2}\cdot x^{-1}-\frac{3}{8} \cdot x^{-2} - ...\right] = x \cdot [ -\frac{1}{6} x^{-1} + ... ] = -\frac{1}{6} + o(x^{-1})$$
which has limit of $-\frac{1}{6}$.
A: Using the degree-one Taylor polynomials for $e^x$ and $\log(1+x)$,
$$  (x^3+x^2+1)^{1/3} - (x^2+x+1)^{0.5} = x \left[ \left(1+\frac1x+\frac1{x^3}\right)^{1/3} - \left(1+\frac1x+\frac1{x^2}\right)^{1/2}\right]= x \left[ e^{\frac13\,\log\left(1+\frac1x+\frac1{x^3}\right)} - e^{\frac12\,\log\left(1+\frac1x+\frac1{x^2}\right)}\right]=x\left[e^{\frac13\,\left(\frac1x+O(\frac1{x^2})\right)}-e^{\frac12\,\left(\frac1x+O(\frac1{x^2})\right)} \right]\\=x\,\left[1+\frac1{3x}+O(\frac1{x^2})-(1+\frac1{2x}+O(\frac1{x^2}))\right] =-\frac16+O(\frac1x)$$
A: $$\lim_{x\to \infty}(x^3+x^2+1)^\frac13-(x^2+x+1)^\frac12$$
$$=\lim_{x\to \infty}x\{\left(1+\frac{x^2+1}{x^3}\right)^\frac13-\left(1+\frac{x+1}{x^2}\right)^\frac12\}$$
$$=\lim_{x\to \infty}x\{(1+\frac13\frac{x^2+1}{x^3}+\cdots)-(1+\frac12\frac{x+1}{x^2}+\cdots)\}$$ using Binomial Expansion for rational exponent which is legal as $\mid\frac{x^2+1}{x^3}\mid<1$ and $\mid\frac{x+1}{x^2}\mid<1$ for $x\to \infty$
$$=\lim_{x\to \infty}\left(\frac{x^2+1}{3x^2}-\frac{x+1}{2x}\right)=\frac13-\frac12=-\frac16$$
